# Finding the molarity of the combination of two solution reacting with each other

I have a question that gives two concentrations and asks for the mass of $$\ce{HCl}$$ formed by the reaction.

$$\ce{H2SO4 + NaCl ->Na2SO4 + HCl}$$

I have two concentrations:

$$\pu{250 mL}$$ of $$\pu{4.00 M}$$ $$\ce{H2SO4}$$, and $$\pu{250 mL}$$ of $$\pu{1.00 M}$$ $$\ce{NaCl}$$.

Here is the balanced reaction equation:

$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.

However, how do I add those two concentrations? I assume the mixture will be at least $$\pu{500 mL}$$, but how do I add the molarity?

Your working equation is correct. $$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

• Find the amount of substance of protons of each solution via $n=c\cdot V$.

$n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

• What is the limiting agent?

Chlorine

• How many moles of hydrogen chloride can only be formed?

$n(\ce{HCl})=0.25~\mathrm{mol}$

• Calculate the mass of hydrogen chloride via $m = n\cdot M$

$M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$

• Martin, you are a formatting genius. Commented Jun 26, 2014 at 16:04
• @Dissenter As a child I loved playing in a sandbox... so I enjoy that now and again too: Check this out! Commented Jun 27, 2014 at 2:09

You're right; you can't directly add the two concentrations since these concentrations are specific to a certain volume of solvent.

If you want to recalculate concentrations of each system component in a combined solution, you need to find the moles of each system component in this combined solution, and then divide by the total solution volume.

• So does that become 500mL of 2.500 M solution?
– Mark
Commented Jun 26, 2014 at 3:50
• That is not quite correct, as you have a limiting agent present. Commented Jun 26, 2014 at 6:16